given that A is an nxn matrix, how do you show that the characteristic polynomial of A is (-1)^n det(A)?
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given that A is an nxn matrix, how do you show that the characteristic polynomial of A is (-1)^n det(A)?
Doing magic because the above is far from being true. In fact,Quote:
Originally Posted by alexandrabel90
is not even a polynomial of degree n, let alone
the characateristic polynomial of any nxn matrix. Read carefully again the question and repost.
Tonio
PS Perhaps there is something about some free coefficient or stuff...?(Giggle)
the exact question is let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the constant term of the characteristic polynomial is (-1)^n det(A)
You know thatQuote:
Originally Posted by alexandrabel90
, right? And that the constant term of a polynomial
is
and
...right?
yup, i know that..
so lets say, A is similar to an upper triangular matric so P(x) = (x-a_11)....(x-a_nn) = x^n + a_n-1 x^(n-1) +....+a_0 so P(0) = a_0 and then how do i continue?
Friend, what if we just pluggedQuote:
Originally Posted by alexandrabel90
into
oh crap!!! im so blur! thanks